Balls to that. I'm armed with a bad attitude and worse math. By my reckoning:
On any given day, a given prisoner has a .99 probability of NOT visiting the room.
Over a period of N days, a given prisoner has a pow(.99, N) probability of having _NEVER_ visited the room. Therefore, the probability a given prisoner HAS visited the room at least once over a period of N days is (1-pow(.99, N)).
Therefore, the probability that X prisoners have each visited the room at least once over a period of N days is pow((1-pow(.99, N)), X)[1]. If I'm in the room, the question I need to answer is: How likely is it that each of the 99 other prisoners have visited the room at least once in the preceeding N days?
After 1000 days, I'm guessing everyone's visited. Screw you guys, I'm going home.[3]
[1]I know this can't be exactly right, because prisoner visits aren't independent events, but I figure it can't be that wrong, and I want to go home.
[2]Cost is calculated in expected days of life forgone, multiplying (1-P(safe)) by 50 years by 365. Arbitrary, but it gives some idea of how much expected life you gain by waiting longer to roll the dice.
[3]I know this is ducking the intention of the question, but I think the point that death isn't that bad a risk if the probability is low is a legitimate one.
On any given day, a given prisoner has a .99 probability of NOT visiting the room.
Over a period of N days, a given prisoner has a pow(.99, N) probability of having _NEVER_ visited the room. Therefore, the probability a given prisoner HAS visited the room at least once over a period of N days is (1-pow(.99, N)).
Therefore, the probability that X prisoners have each visited the room at least once over a period of N days is pow((1-pow(.99, N)), X)[1]. If I'm in the room, the question I need to answer is: How likely is it that each of the 99 other prisoners have visited the room at least once in the preceeding N days?
Let's visit Mr. Python:
After 1000 days, I'm guessing everyone's visited. Screw you guys, I'm going home.[3][1]I know this can't be exactly right, because prisoner visits aren't independent events, but I figure it can't be that wrong, and I want to go home.
[2]Cost is calculated in expected days of life forgone, multiplying (1-P(safe)) by 50 years by 365. Arbitrary, but it gives some idea of how much expected life you gain by waiting longer to roll the dice.
[3]I know this is ducking the intention of the question, but I think the point that death isn't that bad a risk if the probability is low is a legitimate one.